 Solution

Stokes theorem states that ${\oint }_{C}A\cdot dr={\iint }_S(\nabla \times A)\cdot ndS$, with consistent boundary orientation. Here $\nabla \times A=e_z$, so the circulation around the positively oriented planar curve equals its area. Substituting the given parametrization givesThe curve closes at the origin, passes along the right side to $(0,1)$ and returns along the left side, so it is anticlockwise. ConsequentlyEquivalently its two sides are $x=\pm 2y\sqrt{1-y^2}$, $0\le y\le 1$, whose horizontal width integrates to the same result.